# How do you divide (x^4 + 2x^3 +3x -1 )/ (x^2 + 2)?

Sep 9, 2016

${x}^{4} + 2 {x}^{3} + 3 x - 1 = {x}^{2} + 2 x - 2 - \frac{x - 3}{{x}^{2} + 2}$

#### Explanation:

${x}^{4} + 2 {x}^{3} + 3 x - 1$

= $\left({x}^{2} \times {x}^{2}\right) + \underline{2 {x}^{2}} + 2 {x}^{3} - \underline{2 {x}^{2}} + 3 x - 1$

= ${x}^{2} \left({x}^{2} + 2\right) + {x}^{2} \times 2 x + \underline{2 x \times 2} - 2 {x}^{2} - \underline{4 x} + 3 x - 1$

= ${x}^{2} \left({x}^{2} + 2\right) + 2 x \left({x}^{2} + 2\right) - 2 \left({x}^{2} + 2\right) - 4 x + 3 x + 4 - 1$

= $\left({x}^{2} + 2 x - 2\right) \left({x}^{2} + 2\right) - x + 3$

Hence, $\frac{{x}^{4} + 2 {x}^{3} + 3 x - 1}{{x}^{2} + 2}$

= ${x}^{2} + 2 x - 2 - \frac{x - 3}{{x}^{2} + 2}$

Sep 9, 2016

${x}^{2} + 2 x - 2 - \frac{x - 3}{{x}^{2} + 2}$

#### Explanation:

Using place keepers such as $0 {x}^{2}$ to make alignment and calculations more straightforward.

$\text{ } {x}^{4} + 2 {x}^{3} + 0 {x}^{2} + 3 x - 1$
color(magenta)(x^2)(x^2+2) ->ul(x^4+0x^3+2x^2) larr" Subtract
$\text{ } 0 + 2 {x}^{3} - 2 {x}^{2} + 3 x - 1$
$\textcolor{m a \ge n t a}{2 x} \left({x}^{2} + 2\right) \to \text{ "ul(2x^3+0x^2+4x ) larr" Subtract}$
$\text{ } 0 - 2 {x}^{2} - x - 1$
$\textcolor{m a \ge n t a}{- 2} \left({x}^{2} + 2\right) \to \text{ "ul(-2x^2+0x-4 ) larr" Subtract}$
$\textcolor{m a \ge n t a}{\text{ "0-x+3 larr" Remainder}}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{m a \ge n t a}{{x}^{2} + 2 x - 2 + \left[\frac{- x + 3}{{x}^{2} + 2}\right]}$

${x}^{2} + 2 x - 2 - \frac{x - 3}{{x}^{2} + 2}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If you are not sure about the change in sign for the bracket consider this:

Multiply by (+1) but in the form of $\frac{- 1}{- 1}$

$\frac{- 1}{- 1} \left[\frac{- x + 3}{{x}^{2} + 2}\right]$

$\left(- 1\right) \left[\frac{- x + 3}{- 1} \times \frac{1}{{x}^{2} + 2}\right]$

$\left(- 1\right) \left[\left(+ x - 3\right) \times \frac{1}{{x}^{2} + 2}\right]$

$- \left[\frac{x - 3}{{x}^{2} + 2}\right] \text{ "=" } - \frac{x - 3}{{x}^{2} + 2}$